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The sum of the perimeters of an equilateral triangle and a rectangle is $90 \; \text{cm}.$ The area, $\text{T},$ of the triangle and the area, $\text{R},$ of the rectangle, both in $\text{sq cm},$ satisfy the relationship $\text{R = T}^{2}.$ If the sides of the rectangle are in the ratio $1:3,$ then the length, in cm, of the longer side of the rectangle, is

- $24$
- $27$
- $21$
- $18$

## 1 Answer

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Given that, the sum of the perimeter of an equilateral triangle and a rectangle is $90 \; \text{cm}.$

Let the side of an equilateral triangle be $’x’ \; \text{cm}.$

DIAGRAM

Perimeter of equilateral triangle $ = x + x + x = 3x \; \text{cm}.$

Let the length and width of a rectangle be $’a’ \; \text{cm}$ respectively. $(b>a)$

DIAGRAM

We have $ a : b = 1 : 3 $

$\Rightarrow \frac{a}{b} = \frac{1}{3} = k \; (\text let)$

$\Rightarrow \boxed {a = k, b = 3k} $

Perimeter of rectangle $ = 2 (a+b)$

$ = 2 (k+3k)$

$ = 2 (4k) = 8k $

So, $3x + 8k = 90 \; \longrightarrow (1)$

The area of equilateral triangle $\text{T} = \frac{\sqrt{3}}{4} x^{2} \; \text{sq. cm}$

The area of rectangle $\text{R} = ab$

$ = (k) (3k)$

$ = 3k^{2} \; \text{sq. cm}$

We havw $\text{R} = \text{T}^{2}$

$ \Rightarrow 3k^{2} = \left( \frac{\sqrt{3}}{4} x^{2} \right)^{2} $

$ \Rightarrow 3k^{2} = \frac{(\sqrt{3})^{2}}{16} x^{4} $

$ \Rightarrow k^{2} = \frac{1}{16} x^{4} $

$ \Rightarrow \boxed{ k = \frac{1}{4} x^{2}} $

Put the value of $'k’$ in the equation $(1),$ we get.

$ 3x + 8k = 90 $

$ \Rightarrow 3x + 8 \left( \frac{1}{4} x^{2} \right) = 90 $

$ \Rightarrow 2x^{2} + 3x – 90 = 0 $

$ \Rightarrow 2x^{2} + 15x – 12x – 90 = 0 $

$ \Rightarrow x (2x+15) – 6(2x+15) = 0 $

$ \Rightarrow (2x+15) (x-6) = 0 $

$ \Rightarrow 2x + 15 = 0 (\text{or}) x – 6 = 0 $

$ \Rightarrow \boxed{x = \frac{-15}{2} ; x = 6} $

The side of equilateral triangle can’t be negative.

So, $\boxed{x = 6}$

From the equation $(1),$

$ 3x + 8k = 90 $

$ \Rightarrow 3(6) + 8k = 90 $

$ \Rightarrow 8k = 90 – 18 $

$ \Rightarrow 8k = 72 $

$ \Rightarrow \boxed {k = 9}$

Thus, the longer side of the rectangle $ = 3k = 3(9) = 27 \; \text{cm.}$

$\therefore$ The longer side of the rectangle is $27 \; \text{cm}.$

Correct Answer $: \text{B}$

Let the side of an equilateral triangle be $’x’ \; \text{cm}.$

DIAGRAM

Perimeter of equilateral triangle $ = x + x + x = 3x \; \text{cm}.$

Let the length and width of a rectangle be $’a’ \; \text{cm}$ respectively. $(b>a)$

DIAGRAM

We have $ a : b = 1 : 3 $

$\Rightarrow \frac{a}{b} = \frac{1}{3} = k \; (\text let)$

$\Rightarrow \boxed {a = k, b = 3k} $

Perimeter of rectangle $ = 2 (a+b)$

$ = 2 (k+3k)$

$ = 2 (4k) = 8k $

So, $3x + 8k = 90 \; \longrightarrow (1)$

The area of equilateral triangle $\text{T} = \frac{\sqrt{3}}{4} x^{2} \; \text{sq. cm}$

The area of rectangle $\text{R} = ab$

$ = (k) (3k)$

$ = 3k^{2} \; \text{sq. cm}$

We havw $\text{R} = \text{T}^{2}$

$ \Rightarrow 3k^{2} = \left( \frac{\sqrt{3}}{4} x^{2} \right)^{2} $

$ \Rightarrow 3k^{2} = \frac{(\sqrt{3})^{2}}{16} x^{4} $

$ \Rightarrow k^{2} = \frac{1}{16} x^{4} $

$ \Rightarrow \boxed{ k = \frac{1}{4} x^{2}} $

Put the value of $'k’$ in the equation $(1),$ we get.

$ 3x + 8k = 90 $

$ \Rightarrow 3x + 8 \left( \frac{1}{4} x^{2} \right) = 90 $

$ \Rightarrow 2x^{2} + 3x – 90 = 0 $

$ \Rightarrow 2x^{2} + 15x – 12x – 90 = 0 $

$ \Rightarrow x (2x+15) – 6(2x+15) = 0 $

$ \Rightarrow (2x+15) (x-6) = 0 $

$ \Rightarrow 2x + 15 = 0 (\text{or}) x – 6 = 0 $

$ \Rightarrow \boxed{x = \frac{-15}{2} ; x = 6} $

The side of equilateral triangle can’t be negative.

So, $\boxed{x = 6}$

From the equation $(1),$

$ 3x + 8k = 90 $

$ \Rightarrow 3(6) + 8k = 90 $

$ \Rightarrow 8k = 90 – 18 $

$ \Rightarrow 8k = 72 $

$ \Rightarrow \boxed {k = 9}$

Thus, the longer side of the rectangle $ = 3k = 3(9) = 27 \; \text{cm.}$

$\therefore$ The longer side of the rectangle is $27 \; \text{cm}.$

Correct Answer $: \text{B}$